Proving $G\cong \Bbb{Z}_p$ if $|G|$ is prime.

Proving $G\cong \Bbb{Z}_p$ if $|G|$ is prime.

For a group $G$, if $|G|$ is prime, then I have to prove $G\cong \Bbb{Z}_p$.
Take any element $g\in G$. As $G$ cannot have proper subgroups, and as it
also has finite order, $|g|=p$. Hence, $G=\{1,g,g^2,\dots,g^{p-1}\}$. We
know $\Bbb{Z}_p=\{0,1,2,\dots,p-1\}$. Hence, $f:G\to\Bbb{Z}_p$ defined by
$f(g^r)=r$ for any $r\in\Bbb{Z}_p$ is a homomorphism.
If we can now prove that the kernel of this mapping is $\{1\}$, then we're
done. Clearly, $1$ is the only element that maps to $0$ in the above
homomorphism.
Is the above reasoning correct?
What are some other ways of proving isomorphism between the two groups? Is
proving that the kernel is composed of only one element the only way? I
read somewhere that if provided you have proven $f$ is a homomorphism, and
that $h$ is another mapping from $\Bbb{Z}_p$ to $G$, then proving $fg$ is
injective and $gf$ is surjective is sufficient to prove isomorphism. But
what are the properties of $g$?